Find three terms in GP whose product is 8 and the sum is $\dfrac{26}{3}$. \[\]

Hint: We know that any three term of GP can be expressed as $\dfrac{a}{r},a,ar$ where $a$ is any term and $r$ is the ratio. We have from the question that the product is 8 and sum is 26. We solve $\dfrac{a}{r}\times a\times ar=8$ to get $a$ and solve $\dfrac{a}{r}+a+ar=8$ to get $r$ using the obtained value of $a$. \[\]

Complete step by step answer:
A sequence is defined as the enumerated collection of numbers where repetitions are allowed and the order of the numbers matters. It can also express as a one-one map from the natural numbers set to real numbers. The members of the sequence are called terms. Mathematically, a sequence with infinite terms is written as
\[\left( {{x}_{n}} \right)={{x}_{1}},{{x}_{2}},{{x}_{3}},...\]
If the sequence has finite terms terminated by ${{n}^{\text{th}}}$term then we write the sequence as
\[\left( {{x}_{n}} \right)={{x}_{1}},{{x}_{2}},{{x}_{3}},...{{x}_{n}}\]
Geometric sequence otherwise known as Geometric progression, abbreviate d as GP is a type sequence where the ratio between any two consecutive numbers is constant. If $\left( {{x}_{n}} \right)={{x}_{1}},{{x}_{2}},{{x}_{3}},...$ is an GP, then
\[\dfrac{{{x}_{2}}}{{{x}_{1}}}=\dfrac{{{x}_{3}}}{{{x}_{1}}}...=r...(1)\]
Here ratio between two terms is called common ratio and denoted as $r$. The first term is conventionally denoted as $a$. Then we can express any three terms of the GP as
\[\dfrac{a}{r},a,ar\]
We are given that the product of three terms of the given GP in the question is 8. So we have
\[\begin{align}
  & \dfrac{a}{r}\times a\times ar=8 \\
 & \Rightarrow {{a}^{3}}=8 \\
 & \Rightarrow a=2 \\
\end{align}\]
We are also given that the sum of any three terms is $\dfrac{26}{3}$. So we have,
\[\begin{align}
  & \dfrac{a}{r}+a+ar=\dfrac{26}{3} \\
 & \Rightarrow \dfrac{2}{r}+2+2r=\dfrac{26}{3} \\
 & \Rightarrow \dfrac{2}{r}+2r=\dfrac{26}{3}-2=\dfrac{20}{3} \\
 & \Rightarrow \dfrac{1}{r}+r=\dfrac{10}{3} \\
 & \Rightarrow \dfrac{1+{{r}^{2}}}{r}=\dfrac{10}{3} \\
\end{align}\]
We now cross-multiply and obtain quadratic equation in $r$. We have,
\[\begin{align}
  & 3+3{{r}^{2}}=10r \\
 & \Rightarrow 3{{r}^{2}}-10r+3=0 \\
\end{align}\]
We solve the above quadratic equation by splitting the middle term method. We get,
\[\begin{align}
  & \Rightarrow 3{{r}^{2}}-9r-r+3=0 \\
 & \Rightarrow 3r\left( 3r-3 \right)-r\left( r-3 \right)=0 \\
 & \Rightarrow \left( 3r-3 \right)\left( r-3 \right)=0 \\
\end{align}\]
If we take $3r-3=0\Rightarrow r=1$ and we have the three terms of the GP as
\[\dfrac{a}{r},a,ar=\dfrac{2}{1},2,2\cdot 1=2,2,2\]
Here we see the sum of 3 terms $2+2+2\ne \dfrac{26}{3}$ so we reject$r=1$. If we take $3r-1=0\Rightarrow r=\dfrac{1}{3}$ , we have
\[\dfrac{a}{r},a,ar=\dfrac{2}{\left( \dfrac{1}{3} \right)},2,2\left( \dfrac{1}{3} \right)=6,2,\dfrac{2}{3}\]
So the three numbers are $\dfrac{2}{3},2,6.$

Note:
$\dfrac{a}{r}$ exists because in GP $r$ cannot be zero. We have rejected the complex roots of the equation ${{a}^{3}}=8$ because complex number set is not a ordered set. We can alternatively solve taking three terms $a,b,c$in GP with $\dfrac{b}{a}=\dfrac{c}{b}\Rightarrow {{b}^{2}}=ac$